Show that if \([F: K] < \infty\), then the following conditions are equivalent:
\(F\) is Galois over \(K\)
\(F\) is separable over \(K\) and \(F\) is a splitting field of some polynomial \(f\in K[x]\).
\(F\) is a splitting field over \(K\) of some polynomial \(f\in K[x]\) whose irreducible factors are separable.
Solution
Assume throughout that \([F:K]<\infty\).
1. Condition 1 implies condition 2.
Suppose \(F/K\) is Galois. Then \(F/K\) is separable and normal.
Choose finitely many elements \(\alpha_1,\ldots,\alpha_r\in F\) generating \(F\) over \(K\); such a finite generating set exists because \(F/K\) is finite. Let \(m_i\in K[x]\) be the minimal polynomial of \(\alpha_i\) over \(K\) and set \[ f=\prod_{i=1}^r m_i. \] Normality implies that every \(m_i\) splits over \(F\). Hence \(f\) splits over \(F\). Conversely, any splitting field of \(f\) over \(K\) contains every \(\alpha_i\), and therefore contains \[ K(\alpha_1,\ldots,\alpha_r)=F. \] Thus \(F\) is the splitting field of \(f\) over \(K\). Together with separability, this is condition 2.
2. Condition 2 implies condition 3.
Suppose \(F/K\) is separable and \(F\) is the splitting field of \(f\in K[x]\). Let \(p\in K[x]\) be an irreducible factor of \(f\). Since \(f\) splits over \(F\), the polynomial \(p\) has a root \(\alpha\in F\). Because \(F/K\) is separable, \(\alpha\) is separable over \(K\). Its minimal polynomial over \(K\) is \(p\) up to a nonzero scalar, so \(p\) is separable. Thus every irreducible factor of \(f\) is separable, giving condition 3.
3. Condition 3 implies that \(F/K\) is normal.
Suppose \(F\) is the splitting field over \(K\) of a polynomial \(f\in K[x]\) whose irreducible factors are separable. A splitting field over \(K\) of a polynomial in \(K[x]\) is normal over \(K\): every irreducible polynomial in \(K[x]\) with a root in \(F\) splits over \(F\). Hence \(F/K\) is normal.
4. Condition 3 implies that \(F/K\) is separable.
Write the distinct irreducible factors of \(f\) as \[ p_1,\ldots,p_s. \] By hypothesis each \(p_i\) is separable, so every root of every \(p_i\) is separable over \(K\). The field \(F\) is generated over \(K\) by finitely many such roots. A field generated by finitely many elements separable over \(K\) is a separable extension of \(K\). Therefore \(F/K\) is separable.
5. Condition 3 implies condition 1.
By 3 and 4, \(F/K\) is both normal and separable. Since it is finite by hypothesis, it is Galois.
6. The three conditions are equivalent.
The implications 1, 2, and 5 give \[ 1\Longrightarrow2\Longrightarrow3\Longrightarrow1. \]